Question

a^2x +b^2y=c^2,b^2x +a^2y =d^2​

Answer 1

ANSWER:

Given:

• a^2x + b^2y = c^2
• b^2x + a^2y = d^2

To Find:

• Values of x and y.

Solution:

We are given that,

$\implies a^2x+b^2y=c^2- - - -(1)$

And,

$\implies b^2x+a^2y=d^2- - - -(2)$

Multiplying (1) by a^2 and (2) by b^2,

$\implies a^2(a^2x+b^2y=c^2)$

And,

$\implies b^2(b^2x+a^2y=d^2)$

So,

$\implies a^4x+a^2b^2y=a^2c^2- - - -(3)$

And,

$\implies b^4x+a^2b^2y=b^2d^2- - - -(4)$

Subtracting (4) from (3),

$\implies (a^4x+a^2b^2y)-(b^4x+a^2b^2y)=a^2c^2-b^2d^2$

$\implies (a^4-b^4)x+(a^2b^2-a^2b^2)y=a^2c^2-b^2d^2$

$\implies (a^4-b^4)x+0y=a^2c^2-b^2d^2$

$\implies (a^4-b^4)x=a^2c^2-b^2d^2$

Transposing (a^4 - b^4) to RHS,

$\implies\bf x=\dfrac{a^2c^2-b^2d^2}{a^4-b^4}$

Multiplying (1) by b^2 and (2) by a^2,

$\implies b^2(a^2x+b^2y=c^2)$

And,

$\implies a^2(b^2x+a^2y=d^2)$

So,

$\implies a^2b^2x+b^4y=b^2c^2- - - -(5)$

And,

$\implies a^2b^2x+a^4y=a^2d^2- - - -(6)$

Subtracting (6) from (5),

$\implies (a^2b^2x+b^4y)-(a^2b^2x+a^4y)=b^2c^2-a^2d^2$

$\implies (a^2b^2-a^2b^2)x+(b^4-a^4)y=b^2c^2-a^2d^2$

$\implies 0x+(b^4-a^4)y=b^2c^2-a^2d^2$

$\implies (b^4-a^4)y=b^2c^2-a^2d^2$

Transposing (b^4 - a^4) to RHS,

$\implies y=\dfrac{b^2c^2-a^2d^2}{b^4-a^4}$

$\implies\bf y=\dfrac{a^2d^2-b^2c^2}{a^4-b^4}$

Therefore,

$\implies\bf x=\dfrac{a^2c^2-b^2d^2}{a^4-b^4}\:\:\&\:\:y=\dfrac{a^2d^2-b^2c^2}{a^4-b^4}$